Question

A projectile is thrown with an initial velocity $(a\hat{i}+b\hat{j})m{s}^{-1}$, where $\hat{i}$ and $\hat{j}$ are the unit vectors along horizontal and vertical directions respectively. If the range of projectile is twice the maximum height reached by it, then

• A. $b=\frac{a}{2}$
• B. $b=a$
• C. $b=2a$
• D. $b=4a$

Answer 1

The correct option is C $b=2a$

Range $=\frac{2u_x{u}_y}{g}=\frac{2ab}{g}$

Maximum Height $\left(H\right)=\frac{(u_y{)}^2}{2g}=\frac{b^2}{2g}$

Given, $R=2H$

$\Rightarrow \frac{2ab}{g}=2\left(\frac{b^2}{2g}\right)$

$\Rightarrow b=2a$